Trivial extension of an algebra

Question

Suppose that $A$ is a finite dimensional $k$-algebra. Call $Q=\mathrm{Hom}_k(A,k)$. $Q$ admits an $A$-$A$-bimodule structure in the obvious way. The trivial extension of $A$ is defined as follows:

the underlying vectorspace is $T(A)=A\oplus Q$ and the multiplication is given by

$(a,q)(a^\prime,q^\prime)=(aa^\prime,aq^\prime+qa^\prime)$ for $a,a^\prime\in A$ and $q,q^\prime\in Q$.

I want to prove that $T(A)$ is Frobenius (see here for a definition). If we assume to know that the answer to the question linked above is true then we have to prove that $T(A)$ is selfinjective. Any idea how can I prove it?

Answer 1

Check that $T(A)\cong D(T(A))$, where $D=\operatorname{Hom}_k(-,k)$ as $T(A)$-$T(A)$-bimodules (one calls this property being a "symmetric algebra"). The isomorphism is given by $(a,q)\mapsto ((a',q')\mapsto q(a')+q'(a))$. Check that this is an isomorphism of bimodules. We know that $D$ sends projective right modules (like $T(A)$) to injective left modules. Thus $T(A)$ is injective.

EDIT (to answer the comment):

1. $T(A)$ is a projective $T(A)$-module because it is free (see e.g. Lemma 17.1 here).
2. To prove that $D$ send projective right modules to injective left modules note the following (I follow the arguments of Lemma 18.6 here $$D(\bigoplus_{i\in I}P_i)\cong \operatorname{Hom}(\bigoplus P_i,k)\cong \prod_{i\in I}\operatorname{Hom}(P_i,k)\cong \prod_{i\in I}D(P_i)$$ Now since everything is well-behaved with respect to direct sums, direct summands and direct factors, e.g. a direct products of injectives are injective, etc. We just have to prove that $D(A_A)$ is injective.
3. $D(A_A)$ is injective: We check the universal property of injectives. Let $f:X\to Y$ be a monomorphism and $h:X\to D(A)$ be an arbitrary $A$-homomorphism. Let $e:D(A)\to k$ be the $k$-linear map $f\mapsto f(1)$. Now there is a $k$-linear map $e':Y\to K$ such that $e'f=eh$. Now define $h':Y\to D(A)$ by $h'(y)(a)=e'(ay)$. We have to check that it commutes and is an $A$-homomorphism: $$h'(f(x))(a)=e'(af(x))=e'f(ax)=eh(ax)=h(ax)(1)=(ah(x))(1)=h(x)(a)$$ Hence it commutes. Now $A$-homomorpism: $$h'(by)(a)=e'(aby)=h'(y)(ab)=(bh')(y)(a)$$